Aizu - ALDS1_6_B Partition （快排）

Description

Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q.

In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r).

Your task is to read a sequence A and perform the Partition based on the following pseudocode:

Partition(A, p, r)
1 x = A[r]
2 i = p-1
3 for j = p to r-1
4     do if A[j] <= x
5        then i = i+1
6            exchange A[i] and A[j] 
7 exchange A[i+1] and A[r]
8 return i+1

Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].

Input

The first line of the input includes an integer n, the number of elements in the sequence A.

In the second line, Ai (i = 1,2,...,n), elements of the sequence are given separated by space characters.

Output

Print the sorted sequence. Two contiguous elements of the sequence should be separated by a space character. The element which is selected as the pivot of the partition should be indicated by [ ].

Constrains

• 1 ≤ n ≤ 100,000
• 0 ≤ Ai ≤ 100,000

Sample Input

12
13 19 9 5 12 8 7 4 21 2 6 11

Sample Output

9 5 8 7 4 2 6 [11] 21 13 19 12

---

#include <bits/stdc++.h>

using namespace std;

vector<int> num(100005);

int n;

int partition(int start, int end)
{
    int pivot = num[end];
    int pos = start - 1;
    for (int i = start; i <= end - 1; ++i) {
        if (num[i] <= pivot) {
            ++pos;
            std::swap(num[pos], num[i]);
        }
    }
    std::swap(num[pos + 1], num[end]);
    return pos + 1;
}

int main()
{
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    cin >> n;
    for (int i = 0; i < n; ++i) cin >> num[i];

    int pos = partition(0, n - 1);
    for (int i = 0; i < pos; ++i) cout << num[i] << ' ';
    cout << '[' << num[pos] << ']';
    if (pos != n - 1) cout << ' ';
    for (int i = pos + 1; i < n; ++i) {
        cout << num[i];
        if (i != n - 1) cout << ' ';
    }
    cout << endl;

    return 0;
}

注意输出格式，末尾不能有多余空格。

具体解析书中已经写的很详细了。